A Proving Q on Linear Equations (2 Viewers)

[mreditor16]

Can someone help?

 

[VBN2470]

I guess you could setup an augmented matrix [A|b] and get it to REF. Then examine at the last row and see what all the existing possibilities are for each entry. You will always find that one of the columns will be non-leading and hence one parameter will exist --> solution will be given by a line. If not, then b will be your leading column, implying no solution exists for the system and hence the planes won't intersect i.e. they will end up being parallel. NOTE that the last row will never be the zero row, since the two planes are distinct, otherwise you would be solving for intersections points of two identical planes which is redundant.

 

[VBN2470]

Once you get further through the course, you can then say the two distinct planes are parallel if and only if their normal vectors are parallel, otherwise they must intersect on a given line. Similiar to like in R^2, where if you have two distinct lines, they are parallel if and if only their slopes (gradients) are equal, and if they aren't they will then have to intersect at a point (easier to prove this case than for the case for planes in R^3 since less work is needed, but the whole point of this q is to also test your understanding of elementary Gaussian elimination and solubility of system of equations given a set of imposed conditions).

 

[anomalousdecay]

I guess you could setup an augmented matrix [A|b] and get it to REF. Then examine at the last row and see what all the existing possibilities are for each entry. You will always find that one of the columns will be non-leading and hence one parameter will exist --> solution will be given by a line. If not, then b will be your leading column, implying no solution exists for the system and hence the planes won't intersect i.e. they will end up being parallel. NOTE that the last row will never be the zero row, since the two planes are distinct, otherwise you would be solving for intersections points of two identical planes which is redundant.

Following on from this, after obtaining the augmented matrix in REF it should suggest either a line or no solutions which is where you make your argument as to whether the last row suggests that the value is a parameter, which everything else is dependent on (hence giving a line) or suggesting that the last row is a contradiction so that there is no solutions --> parallel.

Use arbitrary values for constants.

 

[mreditor16]

Thanks so much VBN, but this question was in a section before augmented matrices and the like... Like only the topic of "introduction to linear equations" had been introduced before this Q popped up. so then is there a method that doesn't use matrices (not learnt yet, in terms of this Q) etc?

 

[VBN2470]

Thanks so much VBN, but this question was in a section before augmented matrices and the like... Like only the topic of "introduction to linear equations" had been introduced before this Q popped up. so then is there a method that doesn't use matrices (not learnt yet, in terms of this Q) etc?

Then you probably don't need to use matrices, since it is just a system of two equations in three unknowns, so you could solve it as if you are just solving a set of simultaneous equations and use the given conditions to work from there, since you already know the parametric/Cartesian form of a line in R^3
.

 

[mreditor16]

Then you probably don't need to use matrices, since it is just a system of two equations in three unknowns, so you could solve it as if you are just solving a set of simultaneous equations and use the given conditions to work from there, since you already know the parametric/Cartesian form of a line in R^3
.

Wait what lol? So what are you suggesting as a method?

 

[anomalousdecay]

Thanks so much VBN, but this question was in a section before augmented matrices and the like... Like only the topic of "introduction to linear equations" had been introduced before this Q popped up. so then is there a method that doesn't use matrices (not learnt yet, in terms of this Q) etc?

You can also do it without setting up an augmented matrix. Just solve for the variables and show the relationships as VBN said above.

Another method is to find the "points" of intersection through a relationship (which is kinda what I said above). The relationship you obtained will give all possible solutions sets for the "points" of intersection. Then try to deduce that the relationship suggests either a line or no points of intersection (hence parallel) as the only possibilities.

 

[InteGrand]

Can someone help?

Let the first plane be , and the second plane be , where A,B,C,D do not share a common factor (otherwise we would divide through by that), and similarly for E,F,G,H. Also, as the planes are distinct, E/A, F/B, G/C and H/D cannot all be equal.

Multiply equation of by E, and that of by A, to get: and .

For intersection points, we have these two equations as simultaneous equations. So subtract the two equations to get:

.

No solutions exist (i.e. planes do not intersect) iff the following is true: AND AND . Because otherwise, we would have an LHS which has a non-zero coefficient, in which case would could always solve for y and/or z. In this case, BE = AF, CE = AG, AE = AE (obviously), so the normal to the two distinct planes point in the same direction, so the planes are parallel.

Now, otherwise, we will have NOT all of the following true: AND AND . Then we can always solve for z or y and we get a line (I have to go for a bit, I'll explain further if this isn't resolved).

 

[mreditor16]

You can also do it without setting up an augmented matrix. Just solve for the variables and show the relationships as VBN said above.

Another method is to find the "points" of intersection through a relationship (which is kinda what I said above). The relationship you obtained will give all possible solutions sets for the "points" of intersection. Then try to deduce that the relationship suggests either a line or no points of intersection (hence parallel) as the only possibilities.

Oh okay, thanks so so much guys.

 

[InteGrand]

Another method is to find the "points" of intersection through a relationship (which is kinda what I said above). The relationship you obtained will give all possible solutions sets for the "points" of intersection. Then try to deduce that the relationship suggests either a line or no points of intersection (hence parallel) as the only possibilities.

No points of intersection doesn't necessarily imply parallelism though. It is true in , but for planes in higher dimensions, it's not true in general, so this question seems to be requiring you to prove algebraically they are indeed parallel when there are no solutions in .

 

[anomalousdecay]

No points of intersection doesn't necessarily imply parallelism though. It is true in , but for planes in higher dimensions, it's not true in general, so this question seems to be requiring you to prove algebraically they are indeed parallel when there are no solutions.

Question did ask for in

There are numerous ways to do this question. One could even do this using a dot product (too tedious however).

Your solution above however is a lot simpler, though it isn't as obvious.

 

[InteGrand]

Question did ask for in

There are numerous ways to do this question. One could even do this using a dot product (too tedious however).

Your solution above however is a lot simpler, though it isn't as obvious.

I know, I just don't know whether they allow you to assume that in , no solutions imply parallelism, because otherwise, all they're asking you to prove for the "no solutions" part is that it's possible to have no solutions, which is surely obvious?

 

[mreditor16]

tbh guys I'm not getting your solutions. I am up to here and do not know how to continue -

 

[mreditor16]

like I sorta get where you guys are coming from conceptually, but don't know how to deliver the proof.... :/

 

[VBN2470]

Let the first plane be , and the second plane be , where A,B,C,D do not share a common factor (otherwise we would divide through by that), and similarly for E,F,G,H. Also, as the planes are distinct, E/A, F/B, G/C and H/D cannot all be equal.

Multiply equation of by E, and that of by A, to get: and .

For intersection points, we have these two equations as simultaneous equations. So subtract the two equations to get:

.

No solutions exist (i.e. planes do not intersect) iff the following is true: AND AND . Because otherwise, we would have an LHS which has a non-zero coefficient, in which case would could always solve for y and/or z. In this case, BE = AF, CE = AG, AE = AE (obviously), so the normal to the two distinct planes point in the same direction, so the planes are parallel.

Now, otherwise, we will have NOT all of the following true: AND AND . Then we can always solve for z or y and we get a line (I have to go for a bit, I'll explain further if this isn't resolved).

@ mreditor16, this solution is probably the best one to follow and is most likely how they want you to deliver the proof without further knowledge.

This question will never pop up in any of your class tests (or even final), since it is only trying to extend your understanding a little further by utilising a limited amount of tools as possible (hence a [H] question).

 

[mreditor16]

@ mreditor16, this solution is probably the best one to follow and is most likely how they want you to deliver the proof without further knowledge.

tbh I don't get his solution after the third paragraph :/

 

[InteGrand]

tbh guys I'm not getting your solutions. I am up to here and do not know how to continue -

Once you got to the (4) - (3) line, you could say this:

We have two cases:

Case 1: AND AND (note that these all can't be 0, since then we'd have (3) and (4) being identical, which would mean the planes are identical, which is a contradiction)

Case 2: else.

In Case 1, we definitely have no solutions, because our LHS of the (4) - (3) line is 0 but the RHS is non-zero. Also, Case 1 implies parallelism, due to and (and of course ), hence in equations (3) and (4), the planes have equal normal vectors, so the planes are parallel (note that a plane of the form has a normal vector ).

In Case 2, we have in the (4) - (3) line a linear equation in y and z where not both of the coefficients are 0 (if we had them both 0, we'd need the RHS to be 0 too in Case 2, but we said already we can't have that).

This means we can always solve this equation to find y as a function of z say (or y and z may be independent of each other). Once we have found , we can substitute back into one equation of a plane and find x as a function of z too, and then we will get being something with only one variable in it, namely, z. This means z can be renamed a free parameter , and we will end up as a straight line. If did not depend on (i.e. if the coefficient of the z in the (4) - (3) line were 0), y would just be a constant, and we could use the same method to get x as a function of z, and we'd still have a solution with one parameter, i.e. a line.

(Might be helpful to try with some numerical examples.)

 

[mreditor16]

Integrand and VBN, you are two friggin life savers. Let me trawl through your most recent explanations.

 

[VBN2470]

It is like me asking you solve for x and y the following: 0(x) + 0(y) = 4.. No solution right? So the same applies above, you need to work with these constants to see if there exists parameters for your variables, and if they do exist, then you should get one parameter and express all your other variables in terms of this parameter and write it out in parametric form.. If values do not exist, then this means planes do not intersect, implying they are parallel (in R^3). Don't worry about it too much, once you learn solving systems of equations using matrices etc. this question will be pretty easy to do, and even easier when you learn Vector Geometry later in the course..